$\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR}).$ If BC 12cm, find QR.
Download our app for free and get started
Given: $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR})$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{4}}{\text{1}}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\therefore\frac{\text{BC}^2}{\text{QR}^2}=\frac{\text{4}}{\text{1}}$
$\Rightarrow\text{QR}^2=\frac{12^2}{4}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6$
Hence, QR = 6cm
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{4}}{\text{1}}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\therefore\frac{\text{BC}^2}{\text{QR}^2}=\frac{\text{4}}{\text{1}}$
$\Rightarrow\text{QR}^2=\frac{12^2}{4}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6$
Hence, QR = 6cm
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that.View Solution
$(\text{b}^2-\text{c}^2)=2\text{ax}$ - 2The corresponding sides of two similar triangles ABC and DEF are BC = 9.1cm and EF = 6.5cm. If the rerimeter of $\triangle\text{DEF}$ is 25cm, find the perimeter of $\triangle\text{ABC}.$View Solution
- 3In the given figure, side BC of $\triangle\text{ABC}$ is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX Prove that AO : AX = AF : AB and show that EF || BC.View Solution
- 4Find the length of altitude AD of an isosceles $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units.View Solution
- 5Two triangles DEF and GHK are such that $\angle\text{D}=48^\circ$ and $\angle\text{H}=57^\circ.$ If $\triangle\text{DEF}\sim\triangle\text{GHK}$ then find the measure of $\angle\text{F}.$View Solution
- 6In $\triangle\text{ABC},$ the bisector of $\angle\text{B}$ meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively.View Solution
Show that PR × BQ = QR × BP.
- 7View SolutionIn the given figure, DE || BC such that AD = x cm, DB = (3x + 4)cm, AE = (x + 3)cm and EC = (3x + 19)cm. Find the value of x.
- 8In $\triangle\text{ABC},\text{D}$ is the midpoint of $BC$ and $\text{AE}\perp\text{BC}.$ If $\text{AC}>\text{AB},$ show that.View Solution
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$ - 9View SolutionA 13-m-long ladder reaches a window of a building 12m above the ground. Determine the distance of the foot of the ladder from the building.
- 10View SolutionProve that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.