Question
Four points A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that $\frac{\triangle\text{DBC}}{\triangle\text{ABC}}=\frac{1}{2},$ find x.
✓
Answer
Let A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are the vertices of quadrilateral ABCD. AC and BD are joined
Now, area of $\triangle\text{ABC}$ $=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$ $=\frac{1}{2}[6(5+2)+(-3)(-2-3)+4(3-5)]$ $=\frac{1}{2}[6\times7+(-3)(-5)+4(-2)]$ $=\frac{1}{2}[42+15-8]=\frac{49}{2}$ and area of $\triangle\text{DBC},$ $=\frac{1}{2}[\text{x}(5+2)+(-3)(-2-3\text{x})+4(3\text{x}-5)]$ $=\frac{1}{2}[7\text{x}+6+9\text{x}+12\text{x}-20]$ $=\frac{1}{2}[28\text{x}-14]=14\text{x}-7$ Now, $\frac{\triangle\text{DBC}}{\triangle\text{ABC}}=\frac{14\text{x}-7}{\frac{49}{2}}$ $=\frac{(14\text{x}-7)\times2}{49}$ $=\frac{2(14\text{x}-7)}{49}$ $\therefore\ \Big|\frac{2(14\text{x}-7)}{49}\Big|=\frac{1}{2}$ $\Rightarrow\ 4|14\text{x}-7|=\pm49$ If $56\text{x}-28=49$ $\Rightarrow\ 56\text{x}=49+28=77$ $\Rightarrow\ \text{x}=\frac{77}{56}=\frac{11}{8}$ If $4(14\text{x}-7)=-49$ $\Rightarrow\ 56\text{x}-28=-49$ $\Rightarrow\ 56\text{x}=-49+28=-21$ $\Rightarrow\ \text{x}=\frac{-21}{56}=\frac{-3}{8}$ $\therefore\ \text{x}=\frac{11}{8}$ or $\frac{-3}{8}.$
Now, area of $\triangle\text{ABC}$ $=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$ $=\frac{1}{2}[6(5+2)+(-3)(-2-3)+4(3-5)]$ $=\frac{1}{2}[6\times7+(-3)(-5)+4(-2)]$ $=\frac{1}{2}[42+15-8]=\frac{49}{2}$ and area of $\triangle\text{DBC},$ $=\frac{1}{2}[\text{x}(5+2)+(-3)(-2-3\text{x})+4(3\text{x}-5)]$ $=\frac{1}{2}[7\text{x}+6+9\text{x}+12\text{x}-20]$ $=\frac{1}{2}[28\text{x}-14]=14\text{x}-7$ Now, $\frac{\triangle\text{DBC}}{\triangle\text{ABC}}=\frac{14\text{x}-7}{\frac{49}{2}}$ $=\frac{(14\text{x}-7)\times2}{49}$ $=\frac{2(14\text{x}-7)}{49}$ $\therefore\ \Big|\frac{2(14\text{x}-7)}{49}\Big|=\frac{1}{2}$ $\Rightarrow\ 4|14\text{x}-7|=\pm49$ If $56\text{x}-28=49$ $\Rightarrow\ 56\text{x}=49+28=77$ $\Rightarrow\ \text{x}=\frac{77}{56}=\frac{11}{8}$ If $4(14\text{x}-7)=-49$ $\Rightarrow\ 56\text{x}-28=-49$ $\Rightarrow\ 56\text{x}=-49+28=-21$ $\Rightarrow\ \text{x}=\frac{-21}{56}=\frac{-3}{8}$ $\therefore\ \text{x}=\frac{11}{8}$ or $\frac{-3}{8}.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Draw a right triangle ABC in which AB = 6cm, BC = 8cm and $\angle\text{B}=90^\circ.$ Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
→If the roots of the quadratic equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are equal. Then show that $a=b$ $=c$
→Write the expression $a_n - a_k$ for the $A.P. a, a + d, a + 2d, ...$
Hence, find the common difference of the A.P. for which,
$20^{th}$ term is 10 more than the $18^{th}$ term.
→Hence, find the common difference of the A.P. for which,
$20^{th}$ term is 10 more than the $18^{th}$ term.
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that 2AB = DE and BC = 6cm, find EF.
→Evaluate the following:
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
→$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
A road which is 7m wide surrounds a circular park whose circumference is 352m. Find the area of the road.
Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacitites are 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder.
Find the ratio in which the point $\text{P}\Big(\frac{3}{4},\frac{5}{12}\Big)$ divides the line segment joining the points $\text{A}\Big(\frac{1}{2},\frac{3}{2}\Big)$ and B(2, -5).
→Construct a triangle with sides 5cm, 6cm and 7cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.
→Find the middle term of the A.P. 213, 205, 197, ...., 37.