Maths M.C.Q

1. Isosceles but not congruent.

Solution:

In $\triangle\text{ABC},$

AB = AC

$\angle\text{C}=\angle\text{B}$

So, is an isosceles triangle.

But it is given that,

$\angle\text{B}=\angle\text{Q}$

$\angle\text{C}=\angle\text{P}$

$\angle\text{P}=\angle\text{Q}$

So, is also an isosceles triangle.Therefore both triangle are isosceles but not congruent.

2. AC = DE

Solution:

In $\triangle\text{ABC},$

AB = DF

$\angle\text{A}=\angle\text{D}$

We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.

AC = DE.

2. BA > BD

Solution:

In $\triangle\text{ADC},$ AB > BD.

1. 4cm

Solution:

In $\triangle\text{PQR}$ if $\angle\text{R}=\angle\text{P}$NOw, QR = 4cm, Therefore, PQ = 4cm.

2. $\triangle\text{CBA}\cong\triangle\text{PRQ}$

Solution:

We know that, if is congruent to, then sides of $\triangle\text{RST}$ fall on corresponding equal sides of angles of fall on corresponding equal angles.

Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P.

2. $\text{DF}=5\text{CM}, \angle\text{E}=60^{\circ}$

Solution:

Given $\triangle\text{ABC}=\triangle\text{FDE}$ and AB = 5cm, $∠\text{B} = 40°,\angle\text{A}=80^{\circ}$

Since, $\triangle\text{ABC}=\triangle\text{FDE}$ DF = AB

DF = 5cm

$\angle\text{E}=\angle\text{C}$

$\angle\text{E}=\angle\text{C}=180^{\circ}-(\angle\text{A}+\angle\text{B})$

$\angle\text{E}=180^{\circ}-(80^{\circ}+40^{\circ})$

$\angle\text{E}=60^{\circ}$

2. 50º

Solution:

Given $\triangle\text{ABC}$ such that AB = AC and $\angle\text{B}=50^{\circ}$

3. 50º

Solution:

In $\triangle\text{ABC}$ we have

BC = AB But $\angle\text{B}=80^{\circ}$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$

$=\angle\text{A}+80^{\circ}\angle\text{A}=180^{\circ}$

$=2\angle\text{A}=100^{\circ}$

$=\angle\text{A}=100^{\circ}\div2=50^{\circ}$

2. PQ > PR

Solution:

Given, $\angle\text{R}>\angle\text{Q}$PQ > PR.