9th Class Mathematics Triangles Some Important Results

Some Important Results

Category : 9th Class

Some Important Results

(i) The longer side of a triangle has greater angle opposite to it.

(ii) The greater angle of a triangle has longer side opposite to it.

(iii) Perpendicular line segment is the shortest in length. PM is the shortest line segment from point P to line I.

(iv) The distance between line and points lying on it is always zero.

(v) The sum of any two sides of a triangle is greater than the third side.

(vi) The difference between any two sides of a triangle is always less than its third side.

• Ellipses are not all similar to each other.
• Hyperbolas are not all similar to each other.
• In any triarigle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the Morley triangle.
• A non-planar triangle is a triangle which is not contained in a plane.

• Two triangle are said to be similar if their corresponding angles are equal and their corresponding sides are in proportion.
• If a line is drawn parallel to any one side of a triangle intersecting the other two sides at distinct point, then it divides the two sides in the same ratio, it is known as Thale's Theorem.
• The ratio of area of two similar triangles is equal to the ratio of the square of the corresponding sides.
• In a right angled triangle, the square of hypotenuse is equal to the sum of the square of the other two sides, it is known as Pythagoras Theorem.
• Two triangles are said to be congruent if and only if their corresponding sides and angles are equal.

The quadrilateral which is formed by joining the mid points of a given quadrilateral is always_______.

(a) A parallelogram

(b) A trapezium

(c) A rhombus

(d) A rectangle

(e) None of these

Explanation:

Given:

A quadrilateral LMNO is a quadrilateral in which P, Q, R and S are the mid points of LM, MN, NO, OL respectively.

To Prove:

PQRS is a parallelogram

Construction:

Join LN

Proof:

In$\Delta \text{LMN}$,

$PQ\,|\,\,|\,LN$ and $\text{PQ}=\frac{1}{2}\text{LN}$                         ...,.(i)

[Because P and Q are the mid points of LM and MN respectively]

In $\Delta \text{LNO}$

$SR\,|\,\,|LN$ and $\text{SR}=\frac{1}{2}\text{LN}$             .....(ii)

[Because S and R are the mid - points of LO and ON respectively]

From (i) and (ii), we get

$\text{PQ}=\text{SR}$ and $~\text{PQ}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{SR}$                                           ....(iii)

Similarly we can see that

$\text{PS}=\text{QR}$ and $\text{PS}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{QR}$                                    .....(iv)

From (iii) and (iv), we conclude that quadrilateral PQRS is a parallelogram.

In $\Delta \text{LMN}$ which LP is the bisector of MN and BR is the bisector of LP. Here, points P and R are the points of line segment MN and LN respectively then is equal to.... .

(a) $\frac{2}{3}LN$

(b) $\frac{1}{3}LN$

(c) $\frac{2}{3}LM$

(d) $\frac{1}{3}LM$

(e) None of these

Explanation:

Let the Q is the midpoint of LP

Draw a line $PS\,|\,\,|\,MR$ since P is the midpoint of MN

$\Rightarrow$S is the midpoint of RN

In$\Delta \text{LPN}$,

$QR\,|\,\,|\,\,PS$

$\Rightarrow$ R is the midpoint of $LS$ $\Rightarrow$$LR=RS=SN$ $\Rightarrow$$LR=\frac{1}{3}(LN)$

In the figure given below $\text{ON}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{LM}$

the value of$x$is.....

(a) 5, 11

(b) 5, 8

(c) 8, 5

(d) 18, 5

(e) None of these

In the given $\Delta \text{DEF},\text{ GH }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ EF}$ and if $\frac{DG}{GE}=\frac{3}{5}$ and DF = 5.6 cm then the value of DE is equal to..... .

(a) 2.1cm

(b) 3.36cm

(c) 9.35cm

(d) 2.24cm

(e) None of these

In a trapezium PQRS, PQ||RS. If the diagonal PR and QS intersect at a point 0 such that OP = 6 cm and OR = 8 cm, then the ratio of ar($\Delta \text{POQ}$) and ar ($\Delta ROS$) is equal to:

(a) $\frac{3}{4}$

(b) $\frac{36}{48}$

(c) $\frac{9}{16}$

(d) $\frac{5}{16}$

(e) None of these

If the midpoint of the hypotenuse is joined with the vertex of a right angled triangle where the right angle formed then the length of this line segment is:

(a) $\frac{1}{2}$ of the base

(b) $\frac{1}{2}$ of the perpendicular

(c) $\frac{1}{2}$ of the hypotenuse

(d) $\frac{1}{2}$ of the area of triangle

(e) None of these

Explanation:

Given:

The figure given below represents all the conditions S is the midpoint of PR,

To prove:

Construction:

Produce QS to T; so that QS = ST and Join RT.

Proof:

In$\Delta \text{PSQ}$ and $\Delta \text{RST}$ $\text{PS}=\text{SR}$                      (S is the mid point of PR)

$\text{QS}=\text{ST}$                               (By construction)

$\angle \text{PSQ}=\angle \text{TSR}$   (Vertically opposite angles)

$\therefore$$\Delta \text{PSQ}=\Delta \text{RST}$                  (By S-A-S criteria)

$\therefore$$\text{PQ}=\text{RT}$and$\angle \text{QPR}=\angle \text{PRT}$ (Alternate interior angles)

$\Rightarrow$$\text{RT}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{PQ}$ $\Rightarrow$$\angle \text{TRQ}=\text{9}0{}^\circ$

Now in$\Delta \text{PQR}$and$\Delta \text{TRQ}$

$\text{QR}=\text{QR}$              (Common side)

$\text{PQ}=\text{RT}$               (Already proved)

$\therefore$ $\Delta PQR\cong \Delta TRS$                 (By R-H-S criteria)

$\Rightarrow$$PR=QT$$\Rightarrow$$\frac{1}{2}QT=\frac{1}{2}PR$

In the figure given below:

LP, MQ and NR are the median of $\Delta \text{LMN}$ then which one of the following conditions is correct?

(a) $(\text{LM}+\text{MN}+\text{NP})<(\text{LP}+\text{MQ}+\text{NR})$

(b) $\text{(LM}+\text{MN}+\text{NP})>(\text{NP}+\text{MQ}+\text{NR})$

(c) $\text{(LM}+\text{MN}+\text{NP})>(\text{LP}+\text{MQ}+\text{NR)}$

(d) $\text{(L}{{\text{M}}^{2}}+\text{M}{{\text{N}}^{2}}+\text{N}{{\text{P}}^{2}})>(\text{L}{{\text{P}}^{2}}+\text{M}{{\text{Q}}^{2}}+\text{N}{{\text{R}}^{2}}\text{)}$

(e) None of these

One pair of vertically opposite angles of a quadrilateral are a and b respectively as shown in the figure.

Then which one of the following relations is correct?

(a) $\text{a}+\text{b}=\text{c}+\text{d}$

(b) $\text{a}-\text{b}=\text{c}+\text{d}$

(c) $\text{a}+\text{b}=\text{c}-\text{d}$

(d) $\text{a}+\text{b}+\text{c}+\text{d}=\text{36}0{}^\circ$

(e) None of these

In the figure given below:

$\angle \text{DEX}=\angle \text{YEX},\text{ DE}=\text{EX}$ and $\text{EX}=\text{EF}$ then XY is equal to.....

(a) $\angle \text{DEX}$

(b) DF

(c) $\angle \text{XEF}$

(d) EF

(e) None of these

The line segment PQ and RS intersect each other at point 0 in such a way that OP = OS and OQ = OR then which one of the following options is correct?

(a) PR = SQ and $~\text{PR }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ SQ}$

(b) OP=OQ and $~\text{PR}\bcancel{\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }}\text{SQ}$

(c) RO = OS and$~\text{PR }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ SQ}$

(d) PR = SQ and PR and SQ may or may not be parallel to each other

(e) None of these

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