NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Q1 ) In an isosceles triangle ABC, with AB = AC, the bisectors of \(\angle{B}\) and \(\angle{C}\) intersect each other at O. Join A to O. Show that,
(i) OB = OC
(ii) AO bisects \(\angle{A}\)

Answer :

In \(\triangle{ABC}\), we have

AB = AC ...(Given)
Also, \(\angle{B}\) = \(\angle{C}\) ...(Since corresponding angles of equal sides are equal)

Multipling both sides by \(\frac{1}{2} \) , we get,
(\(\frac{1}{2} \) ) \(\angle{B}\) = (\(\frac{1}{2} \) ) \(\angle{C}\)
\(\therefore \) \(\angle{OBC}\) = \(\angle{OCB}\)

Also, it is given that, OB and OC are bisectors of \(\angle{B}\) and \(\angle{C}\), respectively,

Therefore, \(\angle{OBA}\) and \(\angle{OCA}\)
Therefore, OB = OC
...(\(\because \) corresponding sides of equal angles are equal)

Hence, part (i) is proved.



In \(\triangle{ABO}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)
\(\angle{OBA}\) = \(\angle{OCA}\) ...(from part(i))
OB = OC ...(proved earlier)

Therefore, \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ACO}\) ...(By SAS congruency test)

Thus, \(\angle{BAO}\) = \(\angle{CAO}\) ...(By CPCT)

Hence, AO bisects \(\angle{A}\) is proved.

Q2 ) In the \(\triangle{ABC}\), AD is the perpendicular bisector of BC (see figure).
Show that \(\triangle{ABC}\) is an isosceles triangle in which AB = AC.

Answer :

In \(\triangle{ABC}\) and \(\triangle{ACD}\), we have,

DB = DC ...(given)
\(\angle{ADB}\) = \(\angle{ADC}\) ...(Since, AD is the perpendicular bisector of BC)

and AD is the Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, \(\triangle{ABC}\) is an isosceles triangle.
Hence, proved.

Q3 ) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure).
Show that these altitudes are equal.

Answer :

In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\) ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, \(\angle{A}\) is common angle.
and AB = AC ...(Given)

Therefore, \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, BE = CF ...(By CPCT)
Hence, proved.

Q4 ) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\)
(ii) AB = AC i.e., ABC is an isosceles triangle.

Answer :

In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\)(\(\because BE and CF are perpendicular to sides AC and AB)

\(\angle{BAE}\) = \(\angle{CAF}\) ...(\(\because \(\angle{A}\) is the Common angle)

and BE = CF ...(Given)

\(\therefore \) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS Congruency test)

Thus, AB = AC ...(By CPCT)

\(\therefore \) \(\triangle{ABC}\) is an isosceles triangle.
Hence, proved.

Q5 ) ABC and DBC are isosceles triangles on the same base BC (see figure). Show that \(\angle{ABD}\) = \(\angle{ACD}\).

Answer :

In \(\triangle{ABC}\), we have,

AB = AC ...(\(\because \) \(\triangle{ABC}\) is an isosceles triangle)

\(\therefore \) \(\angle{ABC}\) = \(\angle{ACB}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Similarly, in \(\triangle{DBC}\), we have,

BD = CD ...(\(\because \) \(\triangle{DBC}\) too, is an isosceles triangle)

\(\therefore \) \(\angle{DBC}\) = \(\angle{DCB}\) ...(ii) (\(\because \) angles opposite to equal sides are equal)
Now, On adding, Equations (i) and (ii), we get,

\(\Rightarrow \angle{ABC} + \angle{DBC} = \angle{ACB} + \angle{DCB} \)
\(\therefore \) \(\angle{ABD}\) = \(\angle{ACD}\)
Hence, proved.

Q6 ) \(\triangle{ABC}\) is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure).
Show that \(\angle{BCD}\) is a right angle.

Answer :

In \(\triangle{ABC}\), we have,

AB = AC ...(i)(Given)
\(\angle{ACB}\) = \(\angle{ABC}\) ...(ii)(\(\because \) angles opposite to equal sides are equal)
Also, AB = AD ...(iii)(Given)

From (i) and (iii), AC = AD

Now, in \(\triangle{ADC}\) , we have,

AD = AC ...(proved earlier)
\(\angle{ACD}\) = \(\angle{ADC}\) ...(\(\because \) angles opposite to equal sides are equal)

Also, \(\angle{ACD}\) = \(\angle{BDC}\) ...(iv)(\(\because \angle{ADC} = \angle{BDC}\))

On adding Equations (ii) and (iv), we get,

\(\Rightarrow \angle{ACB} + \angle{ACD} = \angle{ABC} + \angle{BDC} \)
\(\Rightarrow \) \(\angle{BCD}\) = \(\angle{ABC}\) + \(\angle{BDC}\)

Adding \(\angle{BCD}\) on both sides, we have,

\( \Rightarrow \angle{BCD} + \angle{BCD} = \angle{ABC} + \angle{BDC} + \angle{BCD} \)
\(\Rightarrow \) 2 \(\angle{BCD}\) = \(180^\circ\) ...(\(\because \) sum of all angles of a triangle is \(180^\circ\))

Therefore, \(\angle{BCD}\) = \(90^\circ\)
Hence, proved.

Q7 ) ABC is a right angled triangle in which \(\angle{A}\) = \(90^\circ\) and AB = AC, find \(\angle{B}\) and \(\angle{C}\).

Answer :

In \(\triangle{ABC}\), we have,

AB = AC ...(Given)
\(\angle{B}\) = \(\angle{C}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Now, we know that,

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\)
\(\Rightarrow \) \(90^\circ\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(given)
\(\Rightarrow \) \(90^\circ\) + \(\angle{B}\) + \(\angle{B}\) = \(180^\circ\) ...(from(i))
\(\Rightarrow \) 2 \(\angle{B}\) = \(180^\circ\) - \(90^\circ\)
\(\Rightarrow \) 2 \(\angle{B}\) = \(90^\circ\)
\(\Rightarrow \) \(\angle{B}\) = \(45^\circ\)
\(\therefore \) \(\angle{C}\) = \(45^\circ\), too.

Q8 ) Show that the angles of an equilateral triangle are \(60^\circ\) each.

Answer :

Let \(\triangle{ABC}\) be an equilateral triangle, such that

AB = BC = CA (by property)



Now, we have,
AB = AC
\(\angle{B}\) = \(\angle{C}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Similarly,
CB = CA

\(\therefore \) \(\angle{A}\) = \(\angle{B}\) ...(ii)(\(\because \) angles opposite to equal sides are equal)

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(iii)(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equations (i),(ii) and (iii), we have,
\(\angle{A}\) + \(\angle{A}\) + \(\angle{A}\) = \(180^\circ\)
\(\Rightarrow \) 3 \(\angle{A}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{A}\) = \(60^\circ\)
\(\Rightarrow \angle{A} = \angle{B} = \angle{C} = 60^\circ\)
Hence, proved.

Q1 ) \(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\)
(ii)\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\)
(iii) AP bisects \(\angle{A}\) as well as \(\angle{D}\)
(iv) AP is the perpendicular bisector of BC.

Answer :

Given: \(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles having same base BC, such that AB = AC and BD = CD.

Proof:
In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)
BD = CD ...(Given)
and AD is Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SSS Congruency test)
Hence, part (i) is proved.


In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,

AB = AC ...(Given)
Also, \(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))
and AP is the common side.

Therefore, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By SAS congruency test)
Hence, part (ii) is proved.


\(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))
Hence, proved that AP bisects \(\angle{A}\)

Now, \(\angle{ADB}\) = \(\angle{ACD}\) ...(\(\because \)\(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Now, by subtracting from \(180^\circ\), we get,

\(180^\circ\) - \(\angle{ADB}\) = \(180^\circ\) - \(\angle{ACD}\)
\(\Rightarrow \) \(\angle{BDP}\) = \(\angle{ACD}\) ...(By linear pair axiom)

Hence, proved that AP bisects \(\angle{D}\)

Hence, part (iii) is proved.

Now, BP = CP (Since, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\))
and \(\angle{c}\) = \(\angle{d}\) ...(i)
But \(\angle{c}\) + \(\angle{d}\) = \(180^\circ\) ...(Linear pair axiom)
\(\Rightarrow \)\(\angle{c}\) + \(\angle{c}\) = \(180^\circ\) ...(from (i))
\(\Rightarrow \) 2 \(\angle{c}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{c}\) = \(90^\circ\)
\(\Rightarrow \)\(\angle{c}\) = \(\angle{d}\) = \(90^\circ\)

Therefore, AP is the perpendicular bisector of BC.
Hence, proved.

Q2 ) AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects \(\angle{A}\)

Answer :

In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,
AB = AC ...(Given)

\(\angle{ABD}\) = \(\angle{ADC}\) = \(90^\circ\) ...(Since AD is an altitude)
and AD is Common side.
Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By RHS congruency axiom)
Therefore, BD = DC ...(By CPCT)
Thus, we can say that, AD bisects BC.

Also, \(\angle{BAD}\) = \(\angle{CAD}\) ...(By CPCT)
Thus, this shows that AD bisects \(\angle{A}\).
Hence, proved.

Q3 ) The two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\triangle{PQR}\) (see figure). Show that
(i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)
(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)

Answer :

We have, AM is the median of \(\triangle{ABC}\),
So, BM = MC = (\(\frac{1}{2} \) ) BC ...(i)(Since, M Bisects BC)

Similarly, PN is the median of \(\triangle{PQR}\).
So, QN = NR = \(\frac{1}{2} \) QR ...(ii)(Since, N Bisects QR)

Now, we have,
BC = QR ...(Given)
Multiplying both sides by \(\frac{1}{2} \), we get,
\(\frac{1}{2} \) BC = \(\frac{1}{2} \) QR
\(\therefore \) BM = QN ...(iii)(From eq.(i) and (ii))

In \(\triangle{ABM}\) and \(\triangle{PQN}\), we have,
AB = PQ ...(Given)
AM = PN ...(Given)
and BM = QN ...(From Eq.(iii))
\(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\) ...(By SSS congruency test)
Hence, part (i) is proved.

In \(\triangle{ABC}\) and \(\triangle{PQR}\), we have,
AB = PQ ...(Given)
\(\angle{B}\) = \(\angle{Q}\) ...(Since, \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\))
and BC = QR ...(Given)
\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\) ...(By SAS congruency test)
Hence, proved.

Q4 ) BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer :

In \(\triangle{BEC}\) and \(\triangle{CBF}\), we have,
\(\angle{BEC}\) = \(\angle{CFB}\) = \(90^\circ\) ...(Given)
BE = CF ...(Given)
and BC is the Common side.
\(\therefore \) \(\triangle{BEC}\) \(\displaystyle \cong \) \(\triangle{CFB}\) ...(By RHS congruency test)
Thus, EC = FB ...(i)(By CPCT)

In \(\triangle{AEB}\) and \(\triangle{AFC}\), we have,
\(\angle{AEB}\) = \(\angle{AFC}\) = \(90^\circ\) ...(Given)
\(\angle{A}\) is Common angle.
and EB = FC ...(Given)
\(\therefore \) \(\triangle{AEB}\) \(\displaystyle \cong \) \(\triangle{AFC}\) ...(By AAS congruency test)
Thus, AE = AF ...(ii)(By CPCT)

Now, on adding Equation (i) and (ii),
we get,

EC + AE = FB + AF
AC = AB ...(Since, AC = EC + AE and AB = FB + AF)

Hence, it is proved that, \(\triangle{ABC}\) is an isosceles triangle.

Q5 ) ABC is an isosceles triangle with AB = AC. Draw AP perpendicular to BC to show that \(\angle{B}\) = \(\angle{C}\)

Answer :

In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,
AB = AC ...(Given)
AP is the Common side.
\(\angle{ABP}\) = \(\angle{ACP}\) = \(90^\circ\) ...(Since, AP perpendicular to BC)
Therefore, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By RHS congruency test)
Thus, \(\angle{B}\) = \(\angle{C}\) ...(By CPCT)
Hence, proved.

Q1 ) Show that in a right angled triangle, the hypotenuse is the longest side.

Answer :

Let ABC be a right angled triangle, such that \(\angle{ABC}\) = \(90^\circ\).
We know that,
\(\therefore \) \(\angle{ABC}\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\)
...(Since, the sums of all angles of a triangle are \(180^\circ\))
\(\Rightarrow \)\(90^\circ\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\) - \(90^\circ\)
\(\Rightarrow \) \(\angle{BCA}\) + \(\angle{CAB}\) = \(90^\circ\)

Thus, we can say that, both the angles \(\angle{BCA}\) and \(\angle{CAB}\) are acute.
\(\therefore \) \(\angle{BCA}\) < \(90^\circ\)and
\(\therefore \) \(\angle{CAB}\) < \(90^\circ\)
\(\therefore \) \(\angle{BCA}\) < \(\angle{ABC}\) and \(\angle{CAB}\) < \(\angle{ABC}\)
Therefore, we can say,

AB < AC and BC < AC ...(Since, side opposite to greater angle is longer)

Hence, the hypotenuse AC is the longest side is proved.

Q2 ) In figure, sides AB and AC of are extended to points P and Q respectively. Also, \(\angle{PBC}\) < \(\angle{QCB}\). Show that AC > AB.

Answer :

By linear pair axiom, we say that,
\(\angle{ACB}\) + \(\angle{QCB}\) = \(180^\circ\) ...(i)
\(\angle{ABC}\) + \(\angle{PBC}\) = \(180^\circ\) ...(ii)
From Equation (i) and (ii), we get,
\(\angle{ACB}\) + \(\angle{QCB}\) = \(\angle{ABC}\) + \(\angle{PBC}\)
But \(\angle{PBC}\) < \(\angle{QCB}\) ...(given)
\(\therefore \) \(\angle{ABC}\) > \(\angle{ACB}\)
\(\therefore \) AC > AB.
Hence, proved.

Q3 ) In figure, \(\angle{B}\) < \(\angle{A}\) and \(\angle{C}\) < \(\angle{D}\). Show that AD < BC.

Answer :

We have, \(\angle{B}\) < \(\angle{A}\) ...(given)
Also, \(\angle{C}\) < \(\angle{D}\) ...(given)
Since, side opposite to greater angle is longer
we get, AO < BO ...(i)
similarly, OD < OC ...(ii)
On adding, Equation (i) and (ii), we have,
AO + OD < BO + OC
\(\therefore \) AD < BC
Hence, proved.

Q4 ) AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\).

Answer :

Given: ABCD is a quadrilateral having AB as the smallest side and CD as the longest side

To prove: \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\)

Construction: Join A to C and B to D.

Figure:

Proof:
In \(\triangle{ABC}\), we have AB is the smallest side.

As, AB < BC
\(\therefore \) \(\angle{5}\) < \(\angle{1}\) ...(i)
(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{ADC}\),
we have CD is the largest side.

As, AD < CD
\(\therefore \) \(\angle{6}\) < \(\angle{2}\) ...(ii)
(Since, angle opposite to longer side is greater)

On adding Equation (i) and (ii), we get,

\(\Rightarrow \) \(\angle{5}\) + \(\angle{6}\) < \(\angle{1}\) + \(\angle{2}\)
\(\Rightarrow \) \(\angle{C}\) < \(\angle{A}\)
\(\Rightarrow \) \(\angle{A}\) > \(\angle{C}\)

Now, Similarly, \(\triangle{ADB}\), we have AB is the smallest side.
As, AD > AB
\(\therefore \) \(\angle{3}\) > \(\angle{8}\) ...(iii)
(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{BCD}\), we have CD is the largest side.
As, CD > BC
\(\therefore \) \(\angle{4}\) > \(\angle{7}\) ...(iv)
(Since, angle opposite to longer side is greater)

On adding Equation (iii) and (iv), we get,
\(\Rightarrow \) \(\angle{3}\) + \(\angle{4}\) > \(\angle{8}\) + \(\angle{7}\)
\(\Rightarrow \) \(\angle{B}\) > \(\angle{D}\)
Hence, proved.

Q5 ) In figure, PR > PQ and PS bisects \(\angle{QPR}\). Prove that \(\angle{PSR}\) > \(\angle{PSQ}\).

Answer :

In \(\triangle{PQR}\), we have,
PR > PQ ...(given)
\(\therefore \) \(\angle{PQR}\) > \(\angle{PRQ}\) ...(i)
(\(\because \) angle opposite to longer side is greater)
We can also say,
\(\angle{a}\) = \(\angle{b}\) ...(ii)
(\(\because \) PS bisects \(\angle{QPR}\))



On adding Equation (i) and (ii), we get,

\(\angle{PQR}\) + \(\angle{a}\)> \(\angle{PRQ}\) + \(\angle{b}\) ...(iii)
Also, \(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(180^\circ\) ...(iv)
and \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\) = \(180^\circ\) ...(v)
(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equation (iv) and (v), we get,

\(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\)
\(\because \) \(\angle{PQS}\) = \(\angle{PQR}\) and \(\angle{PRS}\) = \(\angle{PRQ}\),
we get,

\(\angle{PQR}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRQ}\) + \(\angle{PSR}\) + \(\angle{b}\)

Rearranging the terms, we get,

\(\angle{PQR}\) + \(\angle{a}\) + \(\angle{PSQ}\) = \(\angle{PRQ}\) + \(\angle{b}\) + \(\angle{PSR}\) ...(vi)

From Equation (iii) and (vi), we get,

\(\angle{PSQ}\) < \(\angle{PSR}\) ...
(\(\because \) side opposite to greater angle is longer)
Hence, proved.

Q6 ) Show that of all line segments drawn from a give point not on it, the perpendicular line segment is the shortest.

Answer :

Given: x is a line and A is a point not lying on x.

C is any point on x other than B.

To prove: AB < AC
Proof:
In \(\triangle{ABC}\), \(\angle{B}\) is the right angle.

Therefore, \(\angle{C}\) is an acute angle.
\(\Rightarrow \) \(\angle{C}\) < \(\angle{B}\)
\(\therefore \) AB < AC
(Since, side opposite to greater angle is longer)

Hence, proved that the perpendicular line segment is the shortest.

Q1 ) ABC is a triangle. Locate a point in the interior of \(\triangle \) ABC which is equidistant from all the vertices of \(\triangle \)ABC.

Answer :

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.



In \(\triangle \)ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of \(\triangle \)ABC.

Q2 ) In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer :

The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.


Here, in \(\triangle \)ABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of \(\triangle \)ABC.

Q3 ) In a huge park, people are concentrated at three points (see figure)

A: where these are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]

Answer :

Let us join A and B, and draw l, the perpendicular bisector of AB.
Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O. The point O is the required point where the ice cream parlour be set up.
Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

Q4 ) Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Answer :

It is an activity.
We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii). \(\therefore \)The Fig. (ii) has more triangles.