Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.
Download our app for free and get started
Given: A right triangle ABC.
Let AB = a, BC = b, AC = C and $\text{B}=\angle\text{90}^\circ$
Equilateral triangle with sides AB = a, BC = b and AC = C are drawn respectively.
To prove: Area of equilateral triangle with side hypotenuse (c) is equal to the area of equilateral triangle with side a and b.
$\text{or}\frac{\sqrt{3}}{4}\text{c}^2=\frac{\sqrt{3}}{4}\text{a}^2=\frac{\sqrt{3}}{4}\text{b}^2$
Proof: In $\triangle\text{ABC},$
$\angle\text{ABC}=90^\circ$ [Given]
$\therefore\text{AC}^2=\text{AB}^2+\text{BC}^2$ [by Pythagoras theorem]
$\Rightarrow\text{c}^2=\text{a}^2+\text{b}^2$
$\Rightarrow\frac{\sqrt{3}}{4}\text{c}^2=\frac{\sqrt{3}}{4}\text{a}^2=\frac{\sqrt{3}}{4}\text{b}^2$
[Multiplying by $\frac{\sqrt{3}}{4}$ to both sides]
(Area of equilateral $\triangle$ with side c) = (Area of equilateral $\triangle$ with side a) + (Area of equilateral $\triangle$ with side b)
Hence, the area of equilateral $\triangle$ With hypotenuse is equal to the sum Hence, proved.
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
- 1PQR is a right triangle right anmgled at Q and $\text{QS}\perp\text{PR}.$ If PQ = 6cm and PS = 4cm, find QS, RS and QR.View Solution
- 2In line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle\text{AEF}=\angle\text{AFE}.$ Prove that $\frac{\text{BD}}{\text{CD}}=\frac{\text{BF}}{\text{CE}}.$View Solution
[Hint: Take point G on AB such that CG || DF]
- 3In a quadrilateral ABCD, $\angle\text{A}+\angle\text{D}=90^\circ.$ Prove that $AC^2 + BD^2 = AD^2 + BC^2.$View Solution
[Hint: Produce AB and DC to meet at E] - 4For going to a city B from city A, there is a route via city C such that $\text{AC}\perp\text{CB},$ AC = 2x km and CB = 2(x + 7)km. It is proposed to construct a 26km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.View Solution
- 5View SolutionA 5m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6m to towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
- 6ABC is a triangle right angled at B and $\text{BD}\perp\text{AC}.$ If AD = 4cm, and CD = 5cm, find BD and AB.View Solution
- 7In $\triangle\text{PQR},\text{PD}\perp\text{QR}$ such that D lies on QR. If PQ = a, PR = b, QD = C and DR = d, prove that (a + b)(a - b)=(a + b)(c - d).View Solution
- 8$\text{l}\parallel\text{m}$ and line segments AB, CD and EF are concurrent at point P. Proved that $\frac{\text{AE}}{\text{BF}}=\frac{\text{AC}}{\text{BD}}=\frac{\text{CE}}{\text{FD}}.$View Solution
- 9View SolutionIf PQRS is a parallelogram and AB || PS, then prove that OC || SR.
- 10View SolutionABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18cm, BQ = 35cm and QC = 15cm, find AD.