Question
In $\triangle ABC , \angle ABC =90^{\circ} . \triangle PAB , \triangle QAC$ and $\triangle RBC$ are the equilateral triangles constructed on sides $A B, A C$ and $B C$ respectively. Prove that : $A (\triangle P A B)+ A (\triangle R B C)= A (\triangle Q A C)$.

Answer

Given: In △ABC, ∠ABC = 90°. Equilateral triangles △PAB, △QAC and △RBC are constructed on sides AB, AC and BC respectively.
To prove: A(△PAB) + A(△RBC) = A(△QAC)
Since △PAB is equilateral,
A(△PAB) = $ \frac{\sqrt{3}}{4} AB^{2} $
Since △RBC is equilateral,
A(△RBC) = $ \frac{\sqrt{3}}{4} BC^{2} $
Since △QAC is equilateral,
A(△QAC) = $ \frac{\sqrt{3}}{4} AC^{2} $
In right angled △ABC at B,
$ AB^{2} + BC^{2} = AC^{2} $ (by Pythagoras theorem)
Multiplying both sides by $ \frac{\sqrt{3}}{4} $,
$ \frac{\sqrt{3}}{4} AB^{2} + \frac{\sqrt{3}}{4} BC^{2} = \frac{\sqrt{3}}{4} AC^{2} $
So,
A(△PAB) + A(△RBC) = A(△QAC).
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

A circus tent is cylindrical to a height of 3m and conical above it. If its base radius is 52.5m and the slant height of the conical portion is 53m find the area of canvas needed to make the tent. $\Big[$Take $\pi=\frac{22}{7}\Big]$
The lengths of the diagonals of a rhobbus are 40cm and 42cm. find the length of each side of the rhombus.
A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Solve the following systems of equations:
$\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2,$
$\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1.$
Find the values of a and b for which the following system of equations has infinitely many solutions:
$2x - (2a + 5)y = 5$
$(2b + 1)x - 9y = 15$
A courier company delivered a parcel from Mumbai to Pune. The customer paid ₹ 531 to the courier company. Now, tax invoice shows ₹ 450 as taxable price, CGST is ₹ 40.50 and SGST is ₹ 40.50 , then find the rate of GST applicable in this transaction.
When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.
In the following figure, ABCD is a square of side 2a, Find the ratio between The circumferences. The areas of the in circle and the circum-circle of the square.
Solve the following quadratic equation.
$\frac{1}{4-P}-\frac{1}{2+P}=\frac{1}{4}$
In the given figure, if $AB || CD,$ find the value of $x.$