Class 7 Maths Chapter 7 Questions and Solutions

Final answer: Triangle $ABC$ is equilateral with $AB=BC=CA=4$ cm.

1. Draw a base segment $AB=4$ cm.
2. With centre $A$ and radius $4$ cm, draw an arc above $AB$.
3. With centre $B$ and radius $4$ cm, draw another arc cutting the first arc at $C$.
4. Join $AC$ and $BC$.
5. Since $C$ lies on the arc centred at $A$, $AC=4$ cm. Since $C$ lies on the arc centred at $B$, $BC=4$ cm.

An equilateral triangle has three equal sides. A compass arc marks all points at a fixed distance from its centre.

Final answer: Triangle $ABC$ is equilateral with $AB=BC=CA=4$ cm.

Final answer: All five triangles can be constructed.

1. For $4,4,6$, check $6<4+4=8$, so the triangle exists. It is isosceles.
2. For $3,4,5$, check $5<3+4=7$, so the triangle exists. It is scalene.
3. For $1,5,5$, check $5<1+5=6$, so the triangle exists. It is isosceles.
4. For $4,6,8$, check $8<4+6=10$, so the triangle exists. It is scalene.
5. For $3.5,3.5,3.5$, all sides are equal, so it is an equilateral triangle.

A triangle can be constructed when the largest side is less than the sum of the other two sides.

Final answer: All five triangles can be constructed.

Final answer: Any triangle made from the centre and two points on the circle is isosceles.

1. Let $O$ be the centre of a circle.
2. Choose any two distinct points $P$ and $Q$ on the circle.
3. Join $OP$, $OQ$ and $PQ$.
4. Both $OP$ and $OQ$ are radii of the same circle.
5. Therefore, $OP=OQ$ and triangle $OPQ$ is isosceles.

Every radius of the same circle has the same length, so two radii can be equal sides of a triangle.

Final answer: Any triangle made from the centre and two points on the circle is isosceles.

Final answer: Intersections with two equal radii give isosceles triangles; when the base also equals the radius, the triangle is equilateral.

1. With two equal circles centred at $A$ and $B$, any intersection point $C$ is at the same radius from $A$ and $B$.
2. So $AC=BC$, and $ABC$ is isosceles.
3. If the common radius is also equal to $AB$, then $AB=BC=CA$.
4. In that special case, $ABC$ is equilateral.
5. With three equal circles centred at $A,B,C$, choose intersection points so that the used distances are radii of the same length.

Equal circles have equal radii. Intersections of equal-radius arcs can create triangles with two or three equal sides.

Final answer: Intersections with two equal radii give isosceles triangles; when the base also equals the radius, the triangle is equilateral.

Final answer: No triangle exists for either set of lengths.

1. For $3,4,8$, the longest side is $8$.
2. The sum of the other two sides is $3+4=7$.
3. Since $8>7$, the triangle inequality fails.
4. For $2,3,6$, the longest side is $6$.
5. The sum of the other two sides is $2+3=5$.
6. Since $6>5$, the triangle inequality fails again.

If the longest side is greater than or equal to the sum of the other two sides, the three lengths cannot form a triangle.

Final answer: No triangle exists for either set of lengths.

Final answer: None of the three sets can be side lengths of a triangle.

1. For $10,10,25$, the longest length is $25$ and $10+10=20$. Since $25>20$, no triangle exists.
2. For $5,10,20$, the longest length is $20$ and $5+10=15$. Since $20>15$, no triangle exists.
3. For $12,20,40$, the longest length is $40$ and $12+20=32$. Since $40>32$, no triangle exists.

Compare the longest length with the sum of the other two lengths.

Final answer: None of the three sets can be side lengths of a triangle.

Final answer: Only the longest side needs the serious check: longest side $<$ sum of the other two sides.

1. Let the lengths be $a\le b\le c$.
2. The comparison $a<b+c$ is true because $b$ and $c$ are positive.
3. The comparison $b<a+c$ is also true because $a$ and $c$ are positive and $c\ge b$ is not a problem.
4. The only comparison that can fail is $c<a+b$.
5. So it is enough to test whether the longest side is less than the sum of the two smaller sides.

When lengths are arranged in increasing order, the smaller lengths are automatically less than sums that include the largest length.

Final answer: Only the longest side needs the serious check: longest side $<$ sum of the other two sides.

Final answer: Possible: $3,4,6$; $5,5,8$; $10,20,25$; $24,26,28$. Not possible: $2,2,5$; $2,4,8$; $10,20,35$.

1. $2,2,5$: $5<2+2$ is false, so no triangle.
2. $3,4,6$: $6<3+4=7$, so a triangle exists.
3. $2,4,8$: $8<2+4$ is false, so no triangle.
4. $5,5,8$: $8<5+5=10$, so a triangle exists.
5. $10,20,25$: $25<10+20=30$, so a triangle exists.
6. $10,20,35$: $35<10+20$ is false, so no triangle.
7. $24,26,28$: $28<24+26=50$, so a triangle exists.

A triangle exists exactly when the longest side is smaller than the sum of the other two sides.

Final answer: Possible: $3,4,6$; $5,5,8$; $10,20,25$; $24,26,28$. Not possible: $2,2,5$; $2,4,8$; $10,20,35$.

Final answer: Touch: $2,3,5$, $4,4,8$, $6,7,13$. No intersection: $2,3,6$, $3,4,8$, $5,5,12$. Use $c<a+b$ as the final test.

1. Touching examples have longest side equal to the sum of the other two. Examples: $2,3,5$; $4,4,8$; $6,7,13$.
2. Non-intersecting examples have longest side greater than the sum of the other two. Examples: $2,3,6$; $3,4,8$; $5,5,12$.
3. Procedure: arrange the three lengths in increasing order.
4. Call them $a\le b\le c$.
5. If $c<a+b$, a triangle exists. If $c\ge a+b$, a triangle does not exist.

For base as the longest length, compare it with the sum of the two smaller radii.

Final answer: Touch: $2,3,5$, $4,4,8$, $6,7,13$. No intersection: $2,3,6$, $3,4,8$, $5,5,12$. Use $c<a+b$ as the final test.

Final answer: Triangles exist for $1,100,100$ and $5,10,12$ only.

1. $1,100,100$: $100<1+100=101$, so a triangle exists.
2. $3,6,9$: $9<3+6=9$ is false because equality is not enough, so no triangle exists.
3. $1,1,5$: $5<1+1=2$ is false, so no triangle exists.
4. $5,10,12$: $12<5+10=15$, so a triangle exists.

Use the strict triangle inequality on the longest side.

Final answer: Triangles exist for $1,100,100$ and $5,10,12$ only.

Final answer: Yes. An equilateral triangle exists for any positive side length.

1. For $50,50,50$, the longest side is $50$.
2. The sum of the other two sides is $50+50=100$.
3. Since $50<100$, the triangle exists.
4. For any positive side length $s$, the check is $s<s+s=2s$.
5. This is true whenever $s>0$.

In an equilateral triangle, all three sides are equal, so the triangle inequality becomes $s<2s$.

Final answer: Yes. An equilateral triangle exists for any positive side length.

Final answer: $99<x<101$; $0<x<10$; $4<x<10$, with sample values as listed.

1. For $1$ and $100$, the third length must satisfy $99<x<101$. Five examples are $99.2,99.5,100,100.5,100.8$.
2. For $5$ and $5$, the third length must satisfy $0<x<10$. Five examples are $1,2,5,8,9$.
3. For $3$ and $7$, the third length must satisfy $4<x<10$. Five examples are $5,6,7,8,9$.

If two sides are $a$ and $b$, the third side $x$ must satisfy $|a-b|<x<a+b$.

Final answer: $99<x<101$; $0<x<10$; $4<x<10$, with sample values as listed.

Final answer: All three included-angle triangles can be constructed.

1. For each case, draw one given side as the base.
2. At one endpoint, construct the given included angle.
3. On the new arm, mark the second given side length.
4. Join the marked point to the other endpoint of the base.
5. Since $75^\circ$, $25^\circ$ and $120^\circ$ are all between $0^\circ$ and $180^\circ$, all three triangles are possible.

Two sides and their included angle determine a triangle when both side lengths are positive and the included angle is between 0 degrees and 180 degrees.

Final answer: All three included-angle triangles can be constructed.

Final answer: All three triangles are possible.

1. For $75^\circ,5$ cm,$75^\circ$, the angle sum is $150^\circ<180^\circ$, so construction is possible.
2. For $25^\circ,3$ cm,$60^\circ$, the angle sum is $85^\circ<180^\circ$, so construction is possible.
3. For $120^\circ,6$ cm,$30^\circ$, the angle sum is $150^\circ<180^\circ$, so construction is possible.
4. Construction method: draw the included side, draw the two given angles at its endpoints, and take the intersection of the two arms as the third vertex.

Two base angles and the included side make a triangle if the two angles have sum less than 180 degrees.

Final answer: All three triangles are possible.

Final answer: Possible test pairs: $(70^\circ,30^\circ)$ and $(90^\circ,85^\circ)$. Not possible: $(35^\circ,150^\circ)$ and $(50^\circ,150^\circ)$.

1. For $30^\circ$, possible partners include $40^\circ,100^\circ$; not possible partners include $150^\circ,170^\circ$.
2. For $70^\circ$, possible partners include $30^\circ,80^\circ$; not possible partners include $110^\circ,130^\circ$.
3. For $54^\circ$, possible partners include $60^\circ,100^\circ$; not possible partners include $126^\circ,150^\circ$.
4. For $144^\circ$, possible partners include $10^\circ,30^\circ$; not possible partners include $36^\circ,50^\circ$.
5. $35^\circ+150^\circ=185^\circ$, not possible. $70^\circ+30^\circ=100^\circ$, possible.
6. $90^\circ+85^\circ=175^\circ$, possible. $50^\circ+150^\circ=200^\circ$, not possible.

Two angles can belong to a triangle exactly when their sum is less than 180 degrees.

Final answer: Possible test pairs: $(70^\circ,30^\circ)$ and $(90^\circ,85^\circ)$. Not possible: $(35^\circ,150^\circ)$ and $(50^\circ,150^\circ)$.

Final answer: $72^\circ,15^\circ,60^\circ,60^\circ$; all angles cannot be $70^\circ$; the third angle is $40^\circ$ when two angles are $70^\circ$.

1. $180^\circ-(36^\circ+72^\circ)=72^\circ$.
2. $180^\circ-(150^\circ+15^\circ)=15^\circ$.
3. $180^\circ-(90^\circ+30^\circ)=60^\circ$.
4. $180^\circ-(75^\circ+45^\circ)=60^\circ$.
5. If all three angles were $70^\circ$, the sum would be $210^\circ$, so it is impossible.
6. If two angles are $70^\circ$ each, the third angle is $180^\circ-140^\circ=40^\circ$.

The sum of the three angles of a triangle is always 180 degrees.

Final answer: $72^\circ,15^\circ,60^\circ,60^\circ$; all angles cannot be $70^\circ$; the third angle is $40^\circ$ when two angles are $70^\circ$.

Final answer: $\angle B=65^\circ$ and $\angle C=65^\circ$.

1. Use $\angle A+\angle B+\angle C=180^\circ$.
2. Given $\angle A=50^\circ$, the remaining sum is $180^\circ-50^\circ=130^\circ$.
3. Since $\angle B=\angle C$, divide $130^\circ$ equally.
4. $\angle B=\angle C=65^\circ$.

The angle sum is 180 degrees, and equal angles share the remaining angle measure equally.

Final answer: $\angle B=65^\circ$ and $\angle C=65^\circ$.

Final answer: $\angle ACD=110^\circ$, and it equals $\angle A+\angle B=50^\circ+60^\circ$.

1. First find $\angle ACB$ using the angle sum property.
2. $50^\circ+60^\circ+\angle ACB=180^\circ$.
3. So $\angle ACB=70^\circ$.
4. $\angle ACB$ and exterior angle $\angle ACD$ form a straight angle.
5. Therefore, $\angle ACD=180^\circ-70^\circ=110^\circ$.

An exterior angle of a triangle equals the sum of the two opposite interior angles.

Final answer: $\angle ACD=110^\circ$, and it equals $\angle A+\angle B=50^\circ+60^\circ$.

Final answer: The constructed perpendicular from $A$ to $BC$ is the altitude.

1. Draw $BC=5$ cm.
2. With centre $B$ and radius $6$ cm, draw an arc.
3. With centre $C$ and radius $5$ cm, draw another arc meeting the first arc at $A$.
4. Join $AB$ and $AC$.
5. Place a set square on $BC$ and slide it until its perpendicular edge passes through $A$.
6. Draw the perpendicular from $A$ to $BC$. This segment is the required altitude.

An altitude is a perpendicular line segment from a vertex to the opposite side or its extension.

Final answer: The constructed perpendicular from $A$ to $BC$ is the altitude.

Final answer: Triangle $TRY$ is constructed, and the perpendicular from $T$ to line $RY$ is the altitude.

1. Draw $RY=4$ cm.
2. At $R$, construct an angle of $140^\circ$ with one arm along $RY$.
3. On the other arm, mark $T$ such that $RT=7$ cm.
4. Join $TY$ to complete triangle $TRY$.
5. To construct the altitude from $T$ to $RY$, align a ruler with $RY$.
6. Slide a set square until its perpendicular edge passes through $T$, then draw the perpendicular to the line $RY$.

Use the side-angle-side construction first, then drop a perpendicular from the opposite vertex to the chosen base line.

Final answer: Triangle $TRY$ is constructed, and the perpendicular from $T$ to line $RY$ is the altitude.

Final answer: Infinitely many different triangles exist.

1. Draw segment $AC=5$ cm.
2. A right angle at $B$ means $\angle ABC=90^\circ$.
3. By the circle-with-diameter idea, every point $B$ on the circle having $AC$ as diameter gives a right angle at $B$.
4. There are infinitely many such points on the circle.
5. So infinitely many right triangles satisfy the measurements.

If $AC$ is the hypotenuse and angle $B$ is 90 degrees, point $B$ can be any point on the circle with diameter $AC$, except $A$ and $C$.

Final answer: Infinitely many different triangles exist.

Final answer: Equilateral right or obtuse triangles are impossible. Isosceles right and isosceles obtuse triangles are possible.

1. In an equilateral triangle, all angles are equal.
2. Since the angle sum is $180^\circ$, each angle is $60^\circ$.
3. So an equilateral triangle cannot be right-angled or obtuse-angled.
4. An isosceles right triangle is possible, for example angles $45^\circ,45^\circ,90^\circ$.
5. An isosceles obtuse triangle is possible, for example angles $40^\circ,40^\circ,100^\circ$.

Equilateral triangles have three equal angles of 60 degrees. Isosceles triangles need only two equal sides.

Final answer: Equilateral right or obtuse triangles are impossible. Isosceles right and isosceles obtuse triangles are possible.