Question

Find the area of quadrilateral ABCD, if AP = 8 cm, CQ = 12 cm, BD = 15 cm​

Answer 1

Answer:

AP=8cm

BD=15cm

CQ=12cm

Step-by-step explanation:

formula 1/2*base *height

Answer 2

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Here in the given figure we can notice that the given quadrilateral is divided into 2 identical right angled triangles with given base and given height. So we can find the area of quadrilateral by adding the area of both triangles.

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So let's start!

GIVEN:

• Height of triangle ABD=8cm
• Height of triangle BCD=12cm
• Common base of both triangles=15cm

TO FIND:

• Area of quadrilateral.

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SOLUTION:

~Area of triangle ABD

⇝ Area of ∆ ABD=½×base×height

⇝ Area of ∆ ABD=½×BD×AP

⇝ Area of ∆ ABD=½×15cm×8cm

⇝ Area of ∆ ABD=15cm×4cm

⇝ Area of ∆ ABD=60cm².

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~Area of triangle BCD

⇝ Area of ∆ BCD=½×base×height

⇝ Area of ∆ BCD=½×BD×CQ

⇝ Area of ∆ BCD=½×15cm×12cm

⇝ Area of ∆ BCD=15cm×6cm

⇝ Area of ∆ BCD=90cm².

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~Area of quadrilateral

Area of quadrilateral ABCD=Ar.(∆ ABD+∆BCD)

⇒ (60cm²+90cm²)

⇒ 150cm²

Hence the required area of quadrilateral is 150cm².

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ADDITIONAL INFORMATION:

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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