Question

find the area of quadrilateral ABCD, whose sides are AB=8cm, BC=15cm, CD=12cm,AD=25cm and angleB =90°. Please guys step by step explanation ​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

Sides of Quadrilateral ABCD

AB=8cm;BC=15cm;CD=12cm and AD=25cm

∠B=90°

To Find:

Area of quadrilateral ABCD

In ∆ABC

${AC}^{2} = {AB}^{2} + {BC}^{2}$

$= > {AC}^{2} = {8}^{2} + {15}^{2}$

$= > {AC}^{2} = 64 + 225 = 289$

$AC = \sqrt{289} = 17cm$

Formula:

$\bold{\boxed{Area \: of \: triangle = \frac{1}{2} \times base \times height}}$

$Area \: of \:∆ ABC= \frac{1}{2} \times 15 \times 8 = 60 {cm}^{2}$

IN ∆ACD

AD=25cm;AC=17cm and CD=12cm

$S = \frac{a + b + c}{2} = \frac{25 + 17 + 12}{2} = \frac{54}{2} = 27$

$\bold{\orange{S=27}}$

Area of∆ ACD=

$\sqrt{S(s- a)(s- b)(s - c)}$

$= > \sqrt{27(27 - 17)(27 - 12)27 - 25)}$

$= > \sqrt{27 \times 10 \times 15 \times 2}$

$= \sqrt{3 \times 3 \times 3 \times 2 \times 5 \times 3 \times 5 \times 2}$

$= 3 \times 3 \times 2 \times 5 = 90 {cm}^{2}$

Area of ∆ACD =$90 {cm}^{2}$

Now,$\bold{Area \:of \:ABCD}$ =$\bold{\pink{∆ABC+∆ACD}}$

$= 60 {cm}^{2} + 90 {cm}^{2}$

$\bold{\red{= 150 {cm}^{2}}}$

Answer 2

CONCEPT

USING AREA OF TRIANGLE

PYTHAGORAS THEOREM

HERONS FORMULA

GIVEN

AB=8CM

BC=15CM

CD=12CM

AD=25CM

ANGLE B = 90°

FIND

AREA OF QUADRILATERAL

SOLUTION

AREA OF  TRIANGLE ABC = 1/2× BASE × HEIGHT

1/2×8×15

60CM²

NOW USING PYTHAGORAS THEREOM

AC²=AB²+BC²

AC²=8²+15²

AC²=64+225

AC²=289

AC=17CM

AB=8CM

BC=15CM

BY HERONS FORMULA

S=(A+B+C)/2

S=17+8+15/2

S=32.5 CM

AREA OF TRIANGLE ADC= √32.5(32.5- 17) × 32.5(32.5-8) × 32.5(32.5- 15)

AREA OF TRIANGLE ADC = √503.75×796.25×568.75

AREA OF TRIANGLE ADC = 15104.03 CM²