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Question Number 172453 by mr W last updated on 27/Jun/22

Commented by infinityaction last updated on 27/Jun/22

$22 ? ? ? ?$
Commented by mr W last updated on 27/Jun/22

$y e s . p l e a s e s h o w y o u r w o r k i n g .$
Commented by infinityaction last updated on 27/Jun/22

Commented by infinityaction last updated on 27/Jun/22

$△ O Q R \sim △ R T Q$ $\frac{6}{8} = {(\frac{Q R}{R T})}^{2} \Rightarrow \frac{Q R}{R T} = \frac{\sqrt{3}}{2}$ $l e t Q R = \sqrt{3} k a n d R T = 2 k$ $t h e n Q T = \sqrt{4 k^{2} - 3 k^{2}} = k$ $i n △ P S Q$ $\tan 30 ° = \frac{\sqrt{3} k}{P Q}$ $P Q = 3 k$ $P T = P Q - T Q = 3 k - k = 2 k$ $n o w △ P Q S \sim △ T Q R$ $\frac{A + 2}{8} = {(\frac{3 k}{\sqrt{3} k})}^{2}$ $A + 2 = 24$ $A = 22$
Commented by mr W last updated on 27/Jun/22

$t h a n k s s i r!$
Commented by infinityaction last updated on 27/Jun/22

$M E T H O D - 2$ $△ O Q R \sim △ R T Q$ $\frac{6}{8} = {(\frac{Q R}{R T})}^{2} \Rightarrow \frac{Q R}{R T} = \frac{\sqrt{3}}{2}$ $l e t Q R = \sqrt{3} k a n d R T = 2 k$ $t h e n Q T = \sqrt{4 k^{2} - 3 k^{2}} = k$ $i n △ P S Q$ $\tan 30 ° = \frac{\sqrt{3} k}{P Q}$ $P Q = 3 k$ $P T = P Q - T Q = 3 k - k = 2 k$ $A = △ P Q S - △ O T Q$ $A = \frac{3 \sqrt{3} k^{2}}{2} - 2$ $A = \frac{3 \sqrt{3} k^{2} - 4}{2}$ $i n △ R T Q$ $8 = \frac{\sqrt{3} k^{2}}{2} \Rightarrow k^{2} = \frac{16}{\sqrt{3}}$ $A = \frac{3 \sqrt{3} \times \frac{16}{\sqrt{3}} - 4}{2}$ $A = \frac{48 - 4}{2} \Rightarrow A = 22$
	Answered by mr W last updated on 27/Jun/22

	Commented by mr W last updated on 27/Jun/22

	$Δ E F B \sim Δ B F C \sim Δ C F D$ ${(\frac{B F}{E F})}^{2} = \frac{[Δ B F C]}{[Δ E F B]} = \frac{6}{2} = 3 \Rightarrow \frac{B F}{E F} = \sqrt{3}$ $\frac{C F}{B F} = \frac{B F}{E F} = \sqrt{3}$ $\frac{C F}{E F} = \frac{C F}{B F} \times \frac{B F}{E F} = \sqrt{3} \times \sqrt{3} = 3$ $\frac{[Δ C F D]}{[Δ E F B]} = {(\frac{C F}{E F})}^{2} = 3^{2} = 9$ $S = [Δ C F D] = 9 \times [Δ E F B] = 9 \times 2 = 18$ $T + 2 = S + 6 = 18 + 6 = 24$ $? = T = 24 - 2 = 22 ✓$
	Commented by infinityaction last updated on 27/Jun/22

	$n i c e s o l u t i o n$
	Commented by mr W last updated on 27/Jun/22

	$i n p r i n c i p l e w e h a d s i m i l a r w a y f o r$ $s o l u t i o n .$
	Commented by infinityaction last updated on 27/Jun/22

	$y e s s i r$
	Commented by Tawa11 last updated on 28/Jun/22

	$Great sir$
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