Question

${ \huge{ \boxed{ \boxed{ \red{ \mathfrak{ answer \: please}}}}}}$
⚫QUESTION NO.3⚫


@Sumedhian☺

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Answer 1

Given:ADC and ABC is a Right Angled ∆

✰ AB=7cm

✰ AC=25cm

✰ DC= 20cm

To Find: Area of Figure ABCD = ∆ ACD +∆ ABC

Solution:

Area of Triangle ACD

½ x 25 x AD

To find AD

By Pythagoras Theorem

AC²=AD²+DC²

therefore,

AC²-DC² = AD²

$\implies$ AD = $\large{\sqrt{AC^{2} - DC^{2}}}$

AD=√ 25² - 20²

AD= √ 625 - 400

AD= √ 225

AD=15CM

So therefore

Area of ∆ ACD = ½ x 20 x 15

$\implies$ 300 /2

$\implies$ $\large{\boxed{150}}$

Area of ∆ ABC

½ x AB x BC

To find BC

By Pythagoras theorem

AB² + BC²= AC²

BC²= AC²- AB²

$\implies$ BC = $\large{\sqrt{AC^{2} - AB^{2}}}$

BC= √ 25² -7²

BC= √ 625 -49

BC= √ 576

BC=24

Area of ∆ ABC = ½ x 7 x 24

$\implies$ 7 x 12

$\implies$ $\large{\boxed{84}}$

So area of Figure ABCD = 150 + 84

$\huge\implies$$\large{\boxed{\boxed{234cm^{2}}}}$

Answer 2

Ye rha question.. dekhooooooooo