5 Marks Questions

Question 15 Marks

The lengths of two sides of a triangle are $12 \ cm$ and $15 \ cm$. Between what two measures should the length of the third side fall?

Answer

Let $x cm$ be the length of the third side.
$\because$ Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
$\therefore$ We should have
$12 +15 > x$
$ \therefore 27 > x$
$ \therefore x < 27$
$15 + x >12$
$ \therefore x >12 – 15$
$ \therefore x > – 3$
$x + 12 >15$
$ \therefore x >15 – 12$
$ \therefore x > 3$
$x > –3 and x > 3$
$ \therefore x > 3$
$ \therefore $ The length of the third side should be any length between $3 cm$ and $27 cm.$

View full question & answer→

Question 25 Marks

$ABCD$ is a quadrilateral. Is $AB + BC + CD + DA < 2(AC + BD) ?$

Answer

In \triangle $OAB,$
$OA + OB > AB. . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (i)$
In $\triangle OBC,$
$OB + OC > BC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
In $\triangle OCD,$
$OC + OD > CD. . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iii)$
In $\triangle OAD,$
$OA + OD > AD . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (iv)$
Adding $(1), (2), (3)$ and $(4)$
$2 (OA + OB + OC + OD) > AB + BC + CD + DA$
$ \therefore AB + BC + CD + DA < 2(OA + OB + OC + OD)$
$ \therefore AB + BC + CD + DA < 2(OA + OC + OB + OD)$
$ \therefore AB + BC + CD + DA < 2(AC + BD)$
Hence proved.

View full question & answer→

Question 35 Marks

$ABCD$ is a quadrilateral. Is $AB + BC + CD + DA > AC + BD?$

Answer

In $\triangle ABC,$
$AB + BC > AC . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (i)$
In $\triangle ACD,$
$CD + DA >AC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
Adding $(i)$ and $(ii).$
$AB + BC + CD + DA > 2AC. . . . (iii)$
In $\triangle ABD$
$AB + DA > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iv)$
In $\triangle BCD$
$BC + CD > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (v)$
Adding $(iv)$ and $(v),$
$AB + BC + CD + DA > 2ABD . . . . (vi)$
Adding $(iii)$ and $(vi),$
$2 [AB + BC + CD + DA] > 2(AC + BD)$
$ \therefore AB + BC + CD + DA > AC + BD.$

View full question & answer→

Question 45 Marks

Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.

Answer

First.
Let's construct an isosceles triangle $\ce{ABC}$ of base $BC = 6 \ cm$ and equal sides $\ce{AB = AC} = 8 \ cm$
Steps of construction:
$i.$ Draw line $BC = 6 \ cm$

$ii.$ We need to make $AB$ and $BC$ as $8 \ cm$
Taking $B$ as center, and opening compass to $8 \ cm$, we draw an $arc.$
Now, taking $C$ as center, opening compass to $8 \ cm$, we draw another $arc$

$iii.$ Where both $arc$ intersect is point $A$
Join $AB$ and $AC$

Now,
We know that Mid point of $BC$ is at $3 \ cm$
Let's call it $D.$

Hence,
$AD$ is the median of isosceles $\triangle ABC$

Now,
When we measure $\angle ADC$ by a protector
the angle is $90^{\circ}$
Which means that $AD \perp BC$
$\therefore AD$ is median and altitude of isosceles $\triangle ABC$

View full question & answer→

MATHS 5 Marks Questions

Take a timed test