Exercise 16A

### Find the distance between the points (i) and (ii) and

(i) $\quad \mathrm{A}(9,3)$ and $\mathrm{B}(15,11)$ The given points are $A(9,3)$ and $B(15,11)$. Then $\left(x_{1}=9, y_{1}=3\right)$ and $\left(x_{2}=15, y_{2}=11\right)$ \$A...

### Use ruler and compasses only for this question. (i) Construct the locus of points inside the triangle which are equidistant from B and C. (ii) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

Steps of construction: Draw line $BC\text{ }=\text{ }6\text{ }cm$ and construct angle $CBX\text{ }=\text{ }{{60}^{o}}$. Cut off $AB\text{ }=\text{ }3.5$. Join $AC$, triangle  $ABC$is the...

### Use ruler and compasses only for this question. (i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60degree (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

Steps of construction: (i) Draw line $BC\text{ }=\text{ }6\text{ }cm$ and construct angle $CBX\text{ }=\text{ }{{60}^{o}}$. Cut off $AB\text{ }=\text{ }3.5$. Join $AC$, triangle  $ABC$is...

### In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that: i) point A is equidistant from all the three sides of the triangle. ii) AM bisects angle LMN.

Construction: $Join\text{ }AM$ Proof: (i) Since,$A$ lies on bisector of $\angle N$ So, $A$is equidistant from $MN\text{ }and\text{ }LN.$ Now again, as $A$ lies on the bisector...

### In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.

Construction: From $P$, draw $PL\bot AB~and~PM\bot BC$ Proof: In $\vartriangle PLB\text{ }and\text{ }\vartriangle PMB$, $\angle PLB\text{ }=\angle PMB\text{ }[Each\text{ }{{90}^{o}}]$...

Steps of Construction: i) Draw a line segment $BC\text{ }=\text{ }6.4\text{ }cm$ ii) At$B$, draw a ray $BX$making an angle of ${{75}^{o}}$ with $BC$and cut off $BA\text{ }=\text{ }5\text{... read more ### Construct a right angled triangle PQR, in which ∠Q = 90degree, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meet PR at point T. Prove that T is equidistant from PQ and QR. Steps of Construction: i) Draw a line segment \[QR\text{ }=\text{ }4.5\text{ }cm$ ii) At $Q,$draw a ray $QX$of an angle of ${{90}^{o}}$ iii) With$centre\text{ }R\text{ }and\text{... read more ### In each of the given figures: PA = PB and QA = QB Prove in each case, that PQ (produced, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points. (i) (ii) SOLUTION: Construction: Join \[PQ$which meets $AB\text{ }in\text{ }D.$ Proof: $As\text{ }P\text{ }is\text{ }equidistant\text{ }from\text{ }A\text{ }and\text{ }B.$ So, $P$lies on the...

Given: In triangle ABC, $AB\text{ }=\text{ }4.2\text{ }cm,\text{ }BC\text{ }=\text{ }6.3\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }5\text{ }cm$ Steps of Construction: i) Draw a line segment...