Question

Find graphically the vertices of the triangle whose sides have equations:-
4x + 3y =16 , y = 4 and 4x - 3y = 16.
What type of triangle is this and also find the area and perimeter of the triangle.​

Answer 1

Answer:

An acute angled triangle.

Area = 12 square units

Perimeter = 11 + $4\sqrt{2}$

Step-by-step explanation:

4x + 3y = 16 ............. equation number i

y = 4 .......................... equation number ii

4x - 3y = 16 .............. equation number iii

From equation i and ii:

4x + 12 = 16

=> x = $\frac{16 - 12}{4}$ = 1

∴ Line i and ii intersect at A(1, 4)

From equation ii and iii:

4x - 12 = 16

=> x = $\frac{16 + 12}{4}$ = 7

∴ Line i and ii intersect at B(7, 4)

Subtracting equation iii from equation i:

4x + 3y - 4x + 3y = 16 - 16

=> y = 0, x = 4

∴ Line i and iii intersect at C(4, 0)

$AB^{2}$ = $(1 - 7)^{2}$ + $(4 - 4)^{2}$ = $6^{2}$

AB = 6

$BC^{2}$ = $(7 - 4)^{2}$ + $(4 - 0)^{2}$ = 9 + 16 = 25 = $5^{2}$

BC = 5

$CA^{2}$ = $(4 - 1)^{2}$ + $(0 - 4)^{2}$ = 32

CA = $4\sqrt{2}$

$AB^{2}$ < $BC^{2}$ + $CA^{2}$

∴ ΔABC is acute angled triangle.

Area = $\frac{1}{2}$ x AB x 4 [∵ AB is parallel to X axis at y = 4 line]

= 12 square units

Perimeter = AB + BC + CA = 6 + 5 + $4\sqrt{2}$ = 11 + $4\sqrt{2}$

Answer 2

Answer:

both hard

jejwlpqlwjsiwhwuwiqoqoqowjhdbrbe