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Answer :

Area of triangle = \(\frac{1}{2} \) × base ×height

\(\therefore \) \(\frac{1}{2} \) ×AB×AC = \(\frac{1}{2} \)×BC×AD

Or, \(\frac{1}{2} \)× 5 ×12 = \(\frac{1}{2} \)×13×AD

Or, AD = \(\frac{60}{13} \) = 4.62 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.4 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.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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