Question

ABCD is a quadrilateral whose diagonal BD is of length 26 cm. The lengths of the perpendiculars drawn from A and C on BD are 14 cm and 8 cm respectively. The area of the quadrilateral in sq. cm. is

• A. 196
• B. 244
• C. 286
• D. 324

Hint:

If a diagonal and the perpendicular distances from the remaining vertices to that diagonal are known, the area of the quadrilateral equals \[ \frac{1}{2}\times d\times (h_1+h_2). \]

Answer 1

The Correct Option is C

Concept: The diagonal \(BD\) divides the quadrilateral \(ABCD\) into two triangles: \[ \triangle ABD \quad \text{and} \quad \triangle CBD \] The area of the quadrilateral is the sum of the areas of these two triangles.

Step 1: Find the area of \(\triangle ABD\).
Given: \[ BD=26\text{ cm} \] Perpendicular distance from \(A\) to \(BD\): \[ h_1=14\text{ cm} \] Using the area formula: \[ \text{Area}=\frac{1}{2}\times \text{base}\times \text{height} \] Therefore, \[ \text{Area of }\triangle ABD =\frac{1}{2}\times 26\times 14 =13\times 14 =182 \text{ cm}^2 \]

Step 2: Find the area of \(\triangle CBD\).
Perpendicular distance from \(C\) to \(BD\): \[ h_2=8\text{ cm} \] Hence, \[ \text{Area of }\triangle CBD =\frac{1}{2}\times 26\times 8 =13\times 8 =104 \text{ cm}^2 \]

Step 3: Find the area of quadrilateral \(ABCD\).
\[ \text{Area of quadrilateral} = 182+104 =286 \text{ cm}^2 \] \[ \boxed{286} \]