### 1st Solution Set on Geometry problems for SSC CGL

Solution to 1st set of geometry problems for SSC CGL. The 10 carefully selected problems are on median of triangles, centroid of triangles and more.

Learn how to solve these geometry problems by using basic geometry concepts and geometry problem solving techniques.

Before you go through the solution, take the timed test if not taken yet from,

**Geometry problems for SSC CGL Set 1.**

### Solution to 10 Geometry problems for SSC CGL set 1 - time to answer was 12 mins

#### Problem 1.

$\triangle ABC$ is an isosceles triangle with $AB=AC$ and $AD$ as the median to base $BC$. If $\angle ABC = 35^0$, the $\angle BAD$ is

- $70^0$
- $35^0$
- $55^0$
- $110^0$

#### Solution 1.

The following isosceles triangle has its $\angle ABC = 35^0$ and AB = AC.

As $AD$ is median to the base $BC$, it bisects the side $BC$ so that, $BD=DC$. In two triangles, $\triangle ABD$ and $\triangle ACD$ with common side $AD$ all three pairs of corresponding sides are equal to each other and so the triangles are congruent. This gives us, $\angle ADB = \angle ADC = 90^0$. And so the $\angle BAD = 180^0 - 35^0 - 90^0 = 55^0$.

**Answer:** Option c: $55^0$.

#### Problem 2.

In isosceles triangle $\triangle FGH$, $FG \lt 3$cm and $GH = 8$cm. Then the correct relation is,

- $GH = FH$
- $GH \lt FH$
- $GF = GH$
- $FH \gt GH$

#### Solution 2.

If $FG \lt 3$cm, it is less than half of the second side $GH=8$cm. It means if in the isosceles triangle, the third side equals the smaller side, the two of them will be smaller than the other side $GH$ which violates the basic condition of formation of a triangle. The basic law of formation of a triangle states,

Sum of lengths of any two sides of a triangle must be larger than the third side.

Thus, the equal sides are the larger sides, that is, $GH = FH = 8$cm.

**Answer:** Option a: $GH = FH$.

#### Problem 3.

The sum of three altitudes of a triangle is,

- equal to the sum of three sides
- twice the sum of sides
- greater than the sum of sides
- less than the sum of sides

#### Solution 3.

The altitudes of the triangle $\triangle ABC$ are $AD$, $BE$ and $CF$. As these are the heights, these are the shortest distances to the opposite sides, that is, a height is lesser in length than both its adjacent sides. To be specific, the length of $AD$ will be less than both the adjacent sides, $AB$ and $AC$. This will be true for the other two altitudes $BE$ and $CF$. So if you add up **One adjacent side corresponding to each height,** the sum of heights will always be less than the sum of three sides.

For example we may add up, $BA$ for $AD$, $AC$ for $CF$ and $CB$ for $BE$.

Or, $(AD + CF + BE) \lt (BA + AC + CB)$,

as, $AD \lt BA$, $CF \lt AC$ and $BE \lt CB$.

**Answer:** Option d: less than the sum of sides.

#### Problem 4.

The length of 3 sides of a triangle are, 6cm, 8cm and 10cm. The length of the median to the greatest side is then,

- 5cm
- 8cm
- 4.8cm
- 6cm

#### Solution 4.

As $6^2 + 8^2 = 10^2$ in the given triangle, it is a right triangle $\triangle ABC$ with base side $AB = 8$cm, height side $AC = 6$cm and hypotenuse $BC = 10$cm. $AD$ is the median drawn from $A$ to center point $D$ of largest side $BC$.

We know that the diameter of a circle subtends an angle of $90^0$ at its periphery and so we may consider the three vertices of the triangle to lie on the periphery of the circumscribing circle with diameter as $BC$ and center at $D$. So, $AD$ will be another radius and will be equal to $BD=5$cm.

**Answer:** a: 5cm.

#### Problem 5.

$O$ and $C$ are the Orthocenter and the Circumcenter of an acute angled triangle $\triangle PQR$ respectively. The points $P$ and $O$ are joined and produced to meet the side $QR$ at $S$. If $\angle QCR = 130^0$ and $\angle PQS = 60^0$ then $\angle RPS$ is,

- $100^0$
- $35^0$
- $30^0$
- $60^0$

#### Solution 5.

The following is the circumscribed triangle that depicts the problem.

The chord $QR$ subtends an angle $\angle QCR=130^0$ at the center that is double the $\angle RPQ$ subtended at the periphery. So, $\angle RPQ=65^0$.

Again in right-angled $\triangle PQS$, $\angle PQS = 60^0$. So, the other angle in the $\triangle PQS$, $\angle QPS = 30^0$.

Finally then, desired $\angle RPS = \angle RPQ - \angle QPS = 65^0 - 30^0 = 35^0$.

Answer: b: $35^0$.

#### Problem 6.

If $I$ is the incenter of $\triangle ABC$, $\angle ABC = 65^0$ and $\angle ACB = 55^0$, the $\angle BIC$ is,

- $110^0$
- $120^0$
- $130^0$
- $140^0$

#### S0lution 6.

The following is the picture that depicts the problem.

In $\triangle BIC$ as $BI$ and $CI$ are angle bisectors,

$\angle BIC = 180^0 - \frac{1}{2}(\angle ABC + \angle ACB)$

$\hspace{13mm}= 180^0 - \frac{1}{2}(65^0 + 55^0)$

$\hspace{13mm}=120^0$.

**Answer:** b: $120^0$.

#### Problem 7.

If the median drawn on the base of a triangle is half its base, the triangle will be,

- acute-angled
- obtuse-angled
- right-angled
- equilateral

#### Solution 7.

The median $AD$ drawn on the base $BC$ of the triangle $\triangle ABC$, is equal to half of $BC$, that is, $AD=BD=DC$. This is the situation where we can consider the point $D$ as the center of a circle with the three points $A$, $B$ and $C$ lying on the periphery of the circle and $AD=BD=CD$ as the radii and $BC$ as the diameter. As the diameter subtends an angle of $90^0$, the triangle is a right angled triangle.

**Answer:** c: right-angled.

#### Problem 8.

In a right angled triangle the product of its two sides equals half of the square of the third side which is the hypotenuse. One of the acute angles must then be,

- $15^0$
- $30^0$
- $45^0$
- $60^0$

#### Solution 8.

The following picture depicts the problem.

By the definition of the problem we have,

$AC\times{AB} = \frac{1}{2}BC^2$,

Or, $2AC\times{AB}=BC^2$.

Again by Pythagoras theorem,

$AC^2 + AB^2 = BC^2$.

Subtracting the first equation from the second we have,

$(AC - AB)^2 = 0$,

Or, $AC = AB$.

It means the right-angled triangle is also an isosceles triangle with angles = $45^0$.

Answer: c: $45^0$

#### Problem 9.

In $\triangle ABC$, two points $D$ and $E$ are taken on the lines $AB$ and $BC$ respectively in such a way that $AC$ is parallel to $DE$. The $\triangle ABC$ and $\triangle DBE$ are then,

- always similar
- always congruent
- similar only if $D$ lies outside the line segment $AB$
- congruent only if $D$ lies outside the line segment $AB$

#### Solution 9.

The picture below depicts the problem.

As $AC || DE$, in two triangles $\triangle ABC$ and $\triangle DBE$, $\angle CAB = \angle EDB$, and $\angle ACB = \angle DEB$. Apex angle being common, all three angles of $\triangle ABC$ equal the corresponding angles in $\triangle DBE$. So the two triangles are **similar** to each other.

As $D$ and $E$ are moved down the sides $BA$ and $BC$ respectively keeping the paralellism of $AC||DE$ undisturbed, the two triangles $\triangle ABC$ and $\triangle DBE$ remain similar to each other even when point $D$ lies outside the line segment $AB$.

Only when $D$ and $E$ coincide with $A$ and $C$ respectively, the two triangles become congruent to each other. But even in this case, the *similarity characteristic holds.*

So similarity holds always, whereas congruency hold only on one occasion.

**Answer:** a: always similar.

#### Problem 10.

$AD$ is a median of $\triangle ABC$ and $O$ is the centroid such that $AO = 10cm$. Length of $OD$ (in cm) is,

- 7
- 5
- 4
- 2

#### Solution 10.

$O$ is the centroid of the triangle $\triangle ABC$ and is the point of intersection of its three medians. As the centroid divides a median in the ratio of 2 : 1 from the vertex, in this case we have, $AO : OD = 2 : 1$. As $AO=10$cm, $OD$ is half of it, that is, 5cm.

**Answer:** b: 5.

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