Question

EXERCISE 7.3
1. Find the area of the triangle whose vertices are:
(0 (3).(-1.0), C.-4)
(ii) 6-3-1).(3.-5).(5,2)
2. In each of the following find the value of x. for which the points are collinear
(7.-2).(5.1.3.1
(1) (S, D. (k. 4).(2.-5)
3. Find the area of the triangle formed by joining the mid-points of the sides of the triansk
whose vertices are (0.1).(2, 1and (0.3). Find the ratio of this area to the area of the
given triangle
Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2) 6-3
5. You have studied in Class IX. (Chapter 9. Example 3) that a median of a triangle divides
it into two triangles of equal areas Verify this result for A ABC whose vertices are
A(4-0), B(3-2) and C5.2).​

Answer 1

Answer:

A (4-0)

Step-by-step explanation:

A is a 5.2belong to the equation of quadrilaterl are all sides are equal

Answer 2

Answer:

$\boxed{1}$

(i) (2,3),(-1,0),(2,-4)

Area of triangle $ABC=\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[2(0)-(4))+(-1)(-4-3)+2(3-0)]$

$⇒\frac{1}{2}[2(4)+(-1)(-7)+2(3)]$

$⇒\frac{1}{2}[8+7+6]$

$⇒\frac{1}{2}[21]$

$⇒10.5$square roots

(ii)(-5,-1),(3,-5),(5,2)

ABC=$\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[(-5)(-5-2)+(2-(-1))+5(-1(-5))]$

$⇒\frac{1}{2}[-5(-7)+3(3)+5(4)]$

$⇒\frac{1}{2}[35+9+20]$

$⇒\frac{1}{2}[64]$

$⇒32$square units

$\boxed{2}$

(i) (7,-2),(5,1),(3,k)

$ΔABC=0$

ABC=$\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[7(1-k)+5(k-(-2))+3(-2-1)]=0$

$⇒7(1-k)+5(k+2)+3(-3)=0$

$⇒7-7k+=5k+10-9=0$

$⇒-7k+5k=-10+9-7$

$⇒-2k=-8$

$⇒k=\frac{-8}{-2}$

$⇒k=4$

(ii) (8,1),(k,-4),(2,-5)

Area of $ΔABC=0$

$\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[8(-4-(-5)+k(-5-1)+2(1-(-4))]=0$

$⇒8(-4+5)+k(-6)+2(1+4)=0$

$⇒8(1)+k(-6)+2(5)=0$

$⇒8-6k+10=0$

$⇒-6k=-10-8$

$⇒-6k= -18$

$⇒k=\frac{-18}{-16}$

$⇒k=3$

$\boxed{4}$

Let the vertices of the quadrilateral be,

$A(-4,-2),B(-3,-5)$

$C(3,-2),D(2,3)$

$Joining AC,$

Therefore 2 triangles formed $ABC & ACD$

Hence, Area of quadrilateral $ABCD =Area of Δ ABC +Area of Δ ADC$

Find area Δ ABC

ABC=$\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[-4(-5-(-2))+(-3)(-2-(-2))+3(-2-(-5))]$

$⇒\frac{1}{2}[-4(-5+2)-3(-2+2)+3(-2+5)]$

$⇒\frac{1}{2}[-4(-3)-3(0)+3(3)]$

$⇒\frac{1}{2}[12-0+9]$

$⇒\frac{1}{2}[21]$ square units

Find area Δ ADC,

ADC=$\frac{1}{2}[x_{1}(y_{2})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$⇒\frac{1}{2}[-4(3-(-2))+2(-2(-2))+3(-2-3)]$

$⇒\frac{1}{2}[-4(5)-3(0)+3(-5)]$

$⇒\frac{1}{2}[-20-0-15]$

$⇒\frac{1}{2}[-35]$

But area cannot be negative,

So, Area of the triangle $ADC=\frac{1}{2}[35]$ square units

Hence,

Area of quadrilateral $ABCD=Area of Δ ABC + Area of Δ ABC$

$⇒\frac{21}{2}+\frac{35}{2}$

$⇒\frac{56}{2}$

$⇒28$ Square units

Hope it helps!

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