Question : In the given figure, in $\triangle \operatorname{STU}$, $\mathrm{ST }= 8$ cm, $\mathrm{TU} = 9$ cm and $\mathrm{SU} = 12$ cm. $\mathrm{QU }= 24$ cm, $\mathrm{SR} = 32$ cm and $\mathrm{PT }= 27$ cm. What is the ratio of the area of $\triangle \operatorname{PUQ}$ and the area of $\triangle\operatorname{PTR}$?
Option 1: 1 : 1
Option 2: 4 : 9
Option 3: 2 : 3
Option 4: 5 : 2
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 4 : 9
Solution : Using the Sine rule, In $\triangle \operatorname{PUQ}$, Area of $\triangle \operatorname{PUQ}=\frac{1}{2}× \sin (\operatorname{\angle PUQ})×\operatorname{PU}×\operatorname{QU}$ ⇒ Area of $\triangle \operatorname{PUQ}=\frac{1}{2}× \sin (\operatorname{\angle PUQ})×18×24$ ⇒ Area of $\triangle \operatorname{PUQ}=18× \sin (\operatorname{\angle PUQ}) ×12$ In $\triangle \operatorname{PTR}$, Area of $\triangle \operatorname{PTR}=\frac{1}{2}× \sin (\operatorname{\angle PTR})×\operatorname{PT}×\operatorname{TR}$ ⇒ Area of $\triangle \operatorname{PTR}=\frac{1}{2}× \sin (\operatorname{\angle PTR})×27×24$ ⇒ Area of $\triangle \operatorname{PTR}=27× \sin (\operatorname{\angle PTR})×12$ In $\triangle \operatorname{SUT}$, $\frac{1}{2}× \sin \operatorname{(180^{\circ}–\angle PTR)}×8×9=\frac{1}{2}× \sin \operatorname{(180^{\circ}–\angle PUQ)}×12×9$ ⇒ $4× \sin (\operatorname{\angle PTR})=6× \sin (\operatorname{\angle PUQ})$ ⇒ $\frac{\sin (\operatorname{\angle PTR})}{\sin (\operatorname{\angle PUQ})}=\frac{3}{2}$ $\therefore\frac{\operatorname{Area of \triangle PUQ}}{ \operatorname{Area of \triangle PTR}}=\frac{18× \sin (\operatorname{\angle PUQ})×12}{27× \sin (\operatorname{\angle PTR})×12}=\frac{2×2}{3×3}=\frac{4}{9}$ Hence, the correct answer is 4 : 9.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\triangle ABC \sim \triangle QRP, \frac{\operatorname{area}(\triangle A B C)}{\operatorname{area}(\triangle Q R P)}=\frac{9}{4}, A B=18 \mathrm{~cm}, \mathrm{BC}=15 \mathrm{~cm}$, then the length of $\mathrm{PR}$ is:
Question : The lengths of the three medians of a triangle are $9\;\mathrm{cm}$, $12\;\mathrm{cm}$, and $15\;\mathrm{cm}$. The area (in $\mathrm{cm^2}$) of the triangle is:
Question : The height of an equilateral triangle is $9 \sqrt{3} \mathrm{~cm}$. What is the area of this equilateral triangle?
Question : The radii of two concentric circles are $68\;\mathrm{cm}$ and $22\;\mathrm{cm}$. The area of the closed figure bounded by the boundaries of the circles is ______.
Question : If $\triangle A B C \sim \triangle D E F$, and $B C=4 \mathrm{~cm}, E F=5 \mathrm{~cm}$ and the area of triangle $A B C=80 \mathrm{~cm}^2$, then the area of the $\triangle DEF$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile