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Answer :

(i) Area of the triangle = \(\frac{1}{2} \)× b × h

\(87cm^2 \) = \(\frac{1}{2} \) × 15 × h

Or, h = \(\frac{87×2}{15} \) = \(\frac{174}{15} \) = 11.6cm

(ii) Area of the triangle = \(\frac{1}{2} \)× b × h

\(1256cm^2 \) = \(\frac{1}{2} \) × b × 31.4

Or, b = \(\frac{1256×2}{31.4} \) = 40×2 = 80 mm

(iii) Area of the triangle = \(\frac{1}{2} \)× b × h

\(170.5cm^2 \) = \(\frac{1}{2} \) × 22 × h

Or, h = \(\frac{170.5×2}{22} \) = 15.5 cm

- Q.1 Find the area of each of the following parallelograms:
- Q.2 Find the area of each of the following triangles:
- Q.3 Find the missing values:
- Q.5 PQRS is a parallelogram. QM is the height of Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find: (a) the area of the parallelogram PQRS (b) QN, if PS = 8 cm
- Q.6 DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470\( cm^2 \), AB = 35 cm and AD = 49 cm, find the length of BM and DL.
- Q.7 \(\triangle \) ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of\(\triangle \)ABC. Also find the length of AD.
- Q.8 \(\triangle \)ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ?ABC. What will be the height from C to AB i.e., CE?

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