Question : In $\triangle$PQR, a straight line parallel to the base, QR cuts PQ at X and PR at Y. If PX : XQ = 5 : 6, then XY : QR will be:
Option 1: 5 : 11
Option 2: 6 : 5
Option 3: 11 : 6
Option 4: 11 : 5
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: 5 : 11
Solution : Given: In $\triangle$PQR, QR || XY $\frac{PX}{XQ}=\frac{5}{6}$ Let PX = 5 units, XQ = 6 units So, PQ = 5 + 6 = 11 units In ∆ PXY and ∆ PQR, ∵ QR | | XY ∵ $\angle$X = $\angle$Q ; $\angle$Y = $\angle$R ∴ By AA - similarity, Here, $\triangle$PXY $\simeq \triangle$PQR $\frac{PX}{PQ}=\frac{XY}{QR}$ ⇒ $\frac{5}{11}=\frac{XY}{QR}$ $\therefore XY: QR=5:11$ Hence, the correct answer is 5 : 11.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : A circle is inscribed in $\triangle $PQR touching the sides QR, PR and PQ at the points S, U and T, respectively. PQ = (QR + 5) cm, PQ = (PR + 2) cm. If the perimeter of $\triangle $PQR is 32 cm, then PR is equal to:
Question : $\triangle PQR$ is an isosceles triangle and $PQ=PR=2a$ unit, $QR=a$ unit. Draw $PX \perp QR$, and find the length of $PX$.
Question : In a triangle PQR, $\angle$Q = 90°. If PQ = 12 cm and QR = 5 cm, then what is the radius (in cm) of the circumcircle of the triangle?
Question : $\triangle ABC$ and $\triangle PQR$ are two triangles. AB = PQ = 6 cm, BC = QR =10 cm, and AC = PR = 8 cm. If $\angle ABC = x$, then what is the value of $\angle PRQ$?
Question : In $\triangle$ABC and $\triangle$PQR, AB = PQ and $\angle$B = $\angle$Q. The two triangles are congruent by SAS criteria if:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile