Question

What is the area of the quadrilateral ABCD given below?

• A. $336c{m}^2$
• B. $460c{m}^2$
• C. $542c{m}^2$
• D. $606c{m}^2$

Answer 1

The correct option is A $336c{m}^2$

By drawing the diagonal BD, we can divide the quadrilateral into two triangles – ABD and BCD.​

So, Area of quadrilateral ABCD = Area of triangle BCD + Area of triangle ABD​

Area of triangle ABD = $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times CD\times BC$

= $\frac{1}{2}\times 12\times 16=96c{m}^2$

For triangle BCD, we need to find both base and height. ​

In triangle BCD, ​

Applying Pythagoras theorem,​

$B{D}^2=B{C}^2+C{D}^2$

= ${12}^2+{16}^2$

= 144 + 256 = 400​

​Therefore, BD = 20 cm, which is the base of triangle BCD.​

Now, triangle ABD is an isosceles triangle because AB = AD = 26cm.​

Therefore, a perpendicular from A on BD, which is AE, bisects BD.​

BE = 10 cm​

BD = 2 × BE = 20 cm​

In triangle AEB, ​

Applying Pythagoras theorem,​

$A{B}^2=A{E}^2+B{E}^2$

${26}^2=A{E}^2+{10}^2$

$A{E}^2={26}^2–{10}^2=676–100=576$

AE = 24

Thus, ​

Area of triangle ABD = $\frac{1}{2}\times BD\times AE$

= $\frac{1}{2}\times 20\times 24$ = 240 cm square ​

Hence, Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC ​

= 240 + 96 ​

= 336 cm square​

​Answer: (A) 336 cm square