9th Maths 7.1

NCERT Class 9th solution of Exercise 7.2

NCERT Class 9th solution of Exercise 6.1

NCERT Class 9th Maths Projects

Exercise 7.1

Q1. In quadrilateral `ABCD`, `AC=AD` and `AB` bisects `angleA` (see given figure). Show that `triangleABCcongtriangleABD`. What can you say about `BC` and `BD`?

Sol. :

In `triangleABC` and `triangleABD`

`AC=AD` [Given]

`angleCAB=angleBAD` [`AB` is bisect `angleA`]

`AB=AB` [Common]

`triangleABCcongtriangleABD` [By `SAS` rule]

`BC=BD` [`CPCT`]

Proved.

Q2. `ABCD` is a quadrilateral in which `AD=BC` and `angleDAB=angleCBA` (see given figure). Prove that 

i) `triangleABCcongtriangleBAC` 

ii) `BD=AC` 

iii) `angleABD=angleBAC`.

Sol. ;

i) In `triangleABD`and `triangleBAC`

`AD=BC` [Given]

`angleDAB=angleCBA` [Given]

`AB=AB` [Common]

`triangleABDcongtriangleBAC` [By `SAS` rule]

ii) `triangle ABCcongtriangleBAC` [Proved in (i)]

`BD=AC` [CPCT]

Proved.

iii) `triangleABDcongtriangleBAC` [Proved in (i)]

`angleABD=angleBAC` [CPCT]

Proved.

Q3. `AD` and `BC` are equal perpendicular to a line segment `AB` (see given figure). Show that `CD` bisects `AB`.

Sol. ;

In `triangleAOD` and `triangleBOC`

`BC=AD` [Given]

`angleDAO=angleCBO=90^circ` [Given `BC⊥AB`, `AD⊥AB`]

`angleDOA=angleCOB` [Vertically opposite angles]

`triangleAODcongtriangleBOC` [By `S``A``A` rule]

`AO=OB` [CPCT]

So that `CD` bisect `AB`.

Proved

Q4. `l` and `m` are two parallel lines `p` and `q` (see given figure). Show that `triangleABCcongtriangleCDA`.

Sol. :

In `triangleABC` and `triangleADC`

`l∥m` and `AC` is transversal.

`angleACB=angleDAC` [Alternate angles]

`AC=AC` [Common Side]

`p∥q` and `AC` is transversal.

`angleBAC=angleACD` [Alternate angles]

`triangleABCcongtriangleADC` [By `ASA` rule]

Proved.

Q5. Line`l` is the bisector of an angle `angleA` and `B` is any point on `l`. `BP` and `BQ` are perpendiculars from `B` to the arms of `angleA` (see given figure). Show that:

i) `trangleAPBcongtriangleAQB`

ii) `BP=BQ` or `B` is equidistant from the arms of `angleA`.

Sol. :

i) In `triangleAPB` and `triangleAQB`

`angleBAP=angleBAQ` [Line `l` bisect `angleA`]

`angleAPB=angleAQB` [`BP⊥AP` and `BQ⊥AQ`]

`AB=AB` [Common]

`triangleAPBcongtriangleAOB` [By `S``A``A` rule]

ii) `triangleAPBcongtriangleAQB` [Proved in (i)]

`BP=BQ` [CPCT]

Proved.

Q6. In the given figure `AC=AE, AB=AD` and `angleBAD=angleEAC`. Show that `BC=DE`.

Sol. :

In `triangleABC` and `triangleADE`

`AB=AD`  [Given]

`AC=AE`  [Given]

`angleBAD=angleEAC`  [Given]

We add `angleDAC` both sides]

`angleBAD+angleDAC=angleEAC+angleDAC`

`angleBAC=angleDAE`

`triangleABCcongtriangleADE` [By `SAS` rule]

`BC=DE` [CPCT]

Proved.

Q7. `AB` is a line segment and `p` is its mid-point. `D` and `E` are points on the same side of `AB` such that `angleBAD=angleABE` and `angleEPA=angleDPB` (see given figure). Show that

i) `triangleDAPcongtriangleEBP`

ii) `AD=BE`

Sol. :

i) In `triangleDAP` and `triangleEBP`

`AP=PB`   [`P` is mid point of `AB`]

`angleAPE=angleBPD`  [Given ]

We add `angleEPD` both sides

`angleAPE+angleEPD=angleBPD+angleEPD`

`angleAPD=angleBPE`

`angleDAP=angleEBP` [Given]

`triangleDAPcongtriangleEBP` [By `ASA` rule]

ii) `triangleDPAcongtriangleEBP` [Proved in (i)]

`AD=BE` [CPCT]

Proved.

Q8. In right triangle `ABC`, right angled at `C, M` is the mid-point of hypotenuse `AB`. `C` is joined to `M` and produced to a point `D` such that `DM=CM`. Point `D` is joined to point `B` (see given figure). Show that :

i) `triangleAMCcongtriangleBMD`

ii) `angleDBC` is a right angle`

iii) `triangleDBCcongtriangleACB`

iv) `CM=frac(1)(2)AB`.

Sol. :

i) In `triangleAMC` and `triangleBMD`

`AM=BM` [Given]

`CM=DM` [Given]

`angleCMA=angleBMD` [Vertically opposite angles]

`triangleAMCcongtriangleBMD` [By `SAS` rule]

Proved.

ii) `triangleAMCcongtriangle` [Proved in (i)]

`angleACM=angleBDM` [CPCT]

above angles are Alternate angles, 

so that `DB∥AC`

`angleDBC+angleACB=180^circ` [interior angles of same side]

`angleDBC+90^circ=180^circ`    [`angleACB=90circ`]

`angleDBC=180^circ-90^circ`

So that `angleDCB` is a right angle.

Proved.

iii) `triangleAMCcongtriangleBMD` [Proved in (i)]

`AC=BD` [CPCT]

In `triangleDBC` and `triangleACB`

`BD=AC`   [Proved]

`angleDBC=angleACB=90^circ`  [Proved in (ii)]

`BC=BC`     [Common]

`triangleDBCcongtriangleACB` [By `SAS` rule]

Proved.

iv) `triangleDBCcongtriangleACB`  [Proved in (iii)]

`DC=AB`  [CPCT]

`CM+MD=AB`

`CM+CM=AB` [Given `CM=MD`]

`2CM=AB`

`CM=1/2AB`

Proved.